Prove the trigonometry identity and hence solve given problem

Thanks guys! In summary, the conversation was about the importance of refreshing one's knowledge in trigonometry and the goal of solving 50 questions from different textbooks every day. The conversation also touched on different approaches to solving trigonometric problems, including using Euler's formula and the binomial formula. The conversation ended with the participants sharing and discussing different methods for solving trigonometric equations.
  • #1
chwala
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Homework Statement
Prove that ##\csc ∅ - \cot ∅ = \tan \dfrac {∅}{2}##. Hence, deduce that ##\tan \dfrac {π}{8} = \sqrt{2} -1## and find ##\tan \dfrac {3π}{8}## in terms of ##a+ b\sqrt{2}##
Relevant Equations
Trigonometry.
Refreshing on trig. today...a good day it is...ok find the text problem here; With maths i realize one has to keep on refreshing at all times... my target is to solve 5 questions from a collection of 10 textbooks i.e 50 questions on a day-day basis...motivation from late Erdos :biggrin::wink:...( pure math, applied, stats, mechanics} each and every day starting today...i will only post the interesting questions...

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Ok my approach; let ##\tan \dfrac{∅}{2}=t##

##\dfrac{1}{\sin ∅} - \dfrac{\cos∅ }{\sin ∅}=t##

##\dfrac{1-\cos ∅}{\sin ∅}=t##

##\dfrac{1+t^2-(1-t^2)}{1+t^2} ⋅\dfrac{1+t^2}{2t}=t## thus true.

for second part;

Let ##\dfrac{∅}{2}=\dfrac{π}{8}##

it follows that ##∅ = \dfrac{π}{4}##

therefore; ##\dfrac{1}{\sin \dfrac{1}{4} π} - \dfrac{\cos\dfrac{1}{4}π }{\sin \dfrac{1}{4}π} ##=## \sqrt{2} ## ##-\dfrac {1}{\sqrt{2}}## ⋅## \sqrt{2}##= ##\sqrt{2}-1##

Lastly;

##\tan \dfrac{3π}{8}= \tan \dfrac{π}{4} +\tan \dfrac{π}{8}##

=## \dfrac{\tan \dfrac{π}{4} +\tan \dfrac{π}{8}}{1-\tan \dfrac{π}{4} ⋅\tan \dfrac{π}{8}}##

=##\dfrac{1+\sqrt{2}-1}{1-\sqrt{2}+1} ##

= ##\dfrac{\sqrt{2}}{2-\sqrt{2}}## ⋅ ##\dfrac{2+\sqrt{2}}{2+\sqrt{2}}##

=##\dfrac{2+2\sqrt{2}}{2}= 1+\sqrt{2}##

Of course i know there might be a slightly different approach...would appreciate any comments. Cheers guys!
 
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  • #2
chwala said:
Ok my approach; let ##\tan \dfrac{∅}{2}=t##

##\dfrac{1}{\sin ∅} - \dfrac{\cos∅ }{\sin ∅}=t##
You're assuming what you intend to prove. The only time this is a reasonable approach is when all of the subsequent steps are reversible; i.e., one-to-one operations.

Since you haven't shown that all of your applied operations are one-to-one, a better way is to start with the expression on one side, and then manipulate it to get a sequence of equal expressions that eventually are equal to the other side of the equation you want to prove.
chwala said:
##\dfrac{1-\cos ∅}{\sin ∅}=t##

##\dfrac{1+t^2-(1-t^2)}{1+t^2} ⋅\dfrac{1+t^2}{2t}=t## thus true.
It's not at all obvious to me how the left side above is equal to ##{1 - \cos(\theta)}{\sin(\theta)}##.
If in fact they are equal, then you have proved that t = t, which is trivial.
chwala said:
for second part;

Let ##\dfrac{∅}{2}=\dfrac{π}{8}##

it follows that ##∅ = \dfrac{π}{4}##

therefore; ##\dfrac{1}{\sin \dfrac{1}{4} π} - \dfrac{\cos\dfrac{1}{4}π }{\sin \dfrac{1}{4}π} ##=## \sqrt{2} ## ##-\dfrac {1}{\sqrt{2}}## ⋅## \sqrt{2}##= ##\sqrt{2}-1##

Lastly;

##\tan \dfrac{3π}{8}= \tan \dfrac{π}{4} +\tan \dfrac{π}{8}##

=## \dfrac{\tan \dfrac{π}{4} +\tan \dfrac{π}{8}}{1-\tan \dfrac{π}{4} ⋅\tan \dfrac{π}{8}}##

=##\dfrac{1+\sqrt{2}-1}{1-\sqrt{2}+1} ##

= ##\dfrac{\sqrt{2}}{2-\sqrt{2}}## ⋅ ##\dfrac{2+\sqrt{2}}{2+\sqrt{2}}##

=##\dfrac{2+2\sqrt{2}}{2}= 1+\sqrt{2}##

Of course i know there might be a slightly different approach...would appreciate any comments. Cheers guys!
 
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  • #3
I would start from [tex]
\begin{split}
\cos \theta &= 1 - 2\sin^2(\theta/2) \\
\sin \theta &= 2\cos(\theta/2)\sin(\theta/2) \end{split}[/tex] and substitute these into [tex]
\csc\theta - \cot\theta = \frac{1 - \cos \theta}{\sin \theta}.[/tex]
 
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  • #4
Mark44 said:
You're assuming what you intend to prove. The only time this is a reasonable approach is when all of the subsequent steps are reversible; i.e., one-to-one operations.

Since you haven't shown that all of your applied operations are one-to-one, a better way is to start with the expression on one side, and then manipulate it to get a sequence of equal expressions that eventually are equal to the other side of the equation you want to prove.
It's not at all obvious to me how the left side above is equal to ##{1 - \cos(\theta)}{\sin(\theta)}##.
If in fact they are equal, then you have proved that t = t, which is trivial.
Noted @Mark44 ... Ok we can have;

##\dfrac{1-\cosθ}{\sin θ}=\dfrac{\sin \dfrac{θ}{2}}{cos \dfrac{θ}{2}}##

We know from;

##\cos 2θ =\cos^2θ-\sin^2 θ##

##\cos 2θ =1-2\sin^2 θ## , therefore it follows that;
##\cos θ = 1- 2\sin^2 \dfrac{1}{2}θ##

Also from ##\sin 2θ=2\sin θ \cos θ##, We shall have
##\sin θ = 2 \sin \dfrac{1}{2}θ \cos \dfrac{1}{2}θ##

therefore,

##\dfrac{1-\cosθ}{\sin θ}=\dfrac{1-(1-2\sin^2 \dfrac{1}{2}θ)}{\sin θ}=\dfrac{2\sin^2 \dfrac{1}{2}θ}{2 \sin \dfrac{1}{2}θ \cos \dfrac{1}{2}θ}=\dfrac{\sin \dfrac{θ}{2}}{cos \dfrac{θ}{2}}=\tan \dfrac{θ}{2}##
 
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  • #5
pasmith said:
I would start from [tex]
\begin{split}
\cos \theta &= 1 - 2\sin^2(\theta/2) \\
\sin \theta &= 2\cos(\theta/2)\sin(\theta/2) \end{split}[/tex] and substitute these into [tex]
\csc\theta - \cot\theta = \frac{1 - \cos \theta}{\sin \theta}.[/tex]
Great idea !
 
  • #6
From my university notes; i just noted that we could also use Euler approach and binomal formula for ##\cos 2θ## and ##\sin2θ## ; i.e

##\cos 2θ+ i \sin 2∅= (\cos θ+ i \sin ∅)^2=\cos^2θ+ _2C_1⋅i\cos θ\sinθ+ _2C_2⋅i^2\sin^2 θ##

where;

##\cos 2θ=\cos^2θ+_2C_2⋅i^2\sin^2 θ= \cos^2θ-\sin^2θ## and

##i\sin 2θ=_2C_1⋅i\cos θ\sinθ##

##⇒\sin 2θ=2\sin θ\cos θ##

Bingo!
 
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FAQ: Prove the trigonometry identity and hence solve given problem

What is a trigonometry identity?

A trigonometry identity is an equation that is true for all values of the variables involved. It is used to prove relationships between different trigonometric functions.

How do you prove a trigonometry identity?

To prove a trigonometry identity, you must manipulate the equations using algebraic and trigonometric properties until both sides are equal. This can involve using double angle formulas, sum and difference formulas, and other trigonometric identities.

What is the purpose of solving a trigonometry identity?

Solving a trigonometry identity can help verify the relationships between different trigonometric functions and can also be used to simplify complex expressions involving trigonometric functions.

What are some common strategies for solving trigonometry identities?

Some common strategies for solving trigonometry identities include using basic trigonometric identities, substituting values for the variables, and converting all trigonometric functions to sine and cosine.

How can trigonometry identities be applied to real-world problems?

Trigonometry identities can be applied to real-world problems in fields such as engineering, physics, and astronomy. They can be used to calculate distances, angles, and other measurements in various applications.

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